Question

Tìm giá trị lớn nhất của x đee. X khác 1

E=\(\frac{-x^2+x-10}{x^2-2x+1}\)

Answer 1

\(E=\frac{-x^2+x-10}{x^2-2x+1}=\frac{-\left(x^2-2x+1\right)-x+1-10}{\left(x-1\right)^2}\)

\(=\frac{-\left(x-1\right)^2}{\left(x-1\right)^2}+\frac{-\left(x-1\right)}{\left(x-1\right)^2}-\frac{10}{\left(x-1\right)^2}\)

\(=-1-\frac{1}{x-1}-\frac{10}{\left(x-1\right)^2}\)

Đặt \(t=\frac{1}{x-1}\)

\(\Rightarrow E=-10t^2-t-1=-10\left(t^2+\frac{1}{10}t+\frac{1}{10}\right)=-10\left[\left(t+\frac{1}{20}\right)^2+\frac{39}{400}\right]\)

\(=-10\left(t+\frac{1}{20}\right)^2-\frac{39}{40}\le-\frac{39}{40}\forall t\)

Vậy GTLN của \(E=-\frac{39}{40}\Leftrightarrow t=-\frac{1}{20}\)

                                                \(\Leftrightarrow\frac{1}{x-1}=-\frac{1}{20}\)

                                               \(\Leftrightarrow x-1=-20\)

                                               \(\Leftrightarrow x=-19\)